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⚛️ Physics  ·  Wave Optics  ·  NEET & JEE

In YDSE with glass slabs of thickness t<sub>1</sub> and t<sub>2</sub> and refractive indices n<sub>1</sub> and n<sub>2</sub> placed in front of slits, the shift of central fringe is:

Answer: (n 1 -1)t 1 - (n 2 -1)t 2 divided by λ (in terms of fringes).

  • A (n<sub>1</sub>-1)t<sub>1</sub> - (n<sub>2</sub>-1)t<sub>2</sub> divided by λ (in terms of fringes)
  • B (t<sub>1</sub>-t<sub>2</sub>)/λ, omitting the refractive indices entirely from the path difference
  • C (n<sub>1</sub> t<sub>1</sub> - n<sub>2</sub> t<sub>2</sub>)/λ, omitting the subtraction of unity from each index
  • D (n<sub>1</sub>+n<sub>2</sub>)(t<sub>1</sub>-t<sub>2</sub>)/λ, an incorrect combination of the indices and thicknesses

Correct answer: A. (n<sub>1</sub>-1)t<sub>1</sub> - (n<sub>2</sub>-1)t<sub>2</sub> divided by λ (in terms of fringes)

Explanation: Extra optical path by slab 1: (n<sub>1</sub>-1)t<sub>1</sub>. Slab 2: (n<sub>2</sub>-1)t<sub>2</sub>. Net extra path difference: (n<sub>1</sub>-1)t<sub>1</sub> - (n<sub>2</sub>-1)t<sub>2</sub>. Fringe shift = this net extra path / λ (toward slit 1 if slab 1 has more path).

Young's Double Slit Experimentsourceslits S₁,S₂dscreenbrightdarkPath difference at the screen determines bright (constructive, nλ) vs dark (destructive, (2n-1)λ/2) fringes

Two coherent slits S₁ and S₂ act as secondary sources; at each point on the screen, the path difference between light from S₁ and S₂ determines whether the waves arrive in phase (bright fringe) or out of phase (dark fringe), producing the characteristic alternating fringe pattern.

Concept context

Huygens principle, Young's double slit, diffraction, polarization. Essential for JEE.

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